Let $\Phi$ be a homeomorphism of a compact metric space $M$
which preserves a regular Borel
probability measure $\mu$.(`Regular' $\mu(U) > 0$, if U open. )
Under these hypothesis, I have two questions:

Q1. Is topologically transitivity of $\Phi$
equivalent to ergodicity of $\Phi$?

Q2. If $f$ is a continuous real valued function on $M$ is it true that the pointwise time averages
$f^* (x) = lim_{N \to \infty} \Sigma_1 ^N f(\Phi ^j (x)) /N$ a la Birkhoff exist, not
just for almost every x, but for every darn x?

Motivation. I just went through E. Hopf's proof (as presented in a
1971 BAMS article ) that for compact negatively curved
surfaces the geodesic flow on the unit tangent bundle is ergodic. That proof
gets much simpler if you get rid of the measure-theory/Birkhoff ergodic theorem
business, which is to say, if the answer to flow versions of either Q1 or Q2 is `yes'.

This is a non-standard definition of 'regular', I think. I would use the term 'fully-supported' for the property that every open set has positive measure. In any case the answer is still no: there can be fully-supported non-ergodic measures. For the example in my answer below ($M$ is the full two-shift, $\Phi$ is the shift map), let $\mu_1$ and $\mu_2$ be two distinct Bernoulli measures, and let $\mu=\frac 12(\mu_1+\mu_2)$. Then $\mu$ is invariant but not ergodic, even though $\Phi$ is topologically transitive.
–
Vaughn ClimenhagaNov 22 '12 at 2:24

More generally you can just take $\nu_1\neq \nu_2$ to be any invariant measures. If one of them is fully-supported then $\mu=a\nu_1 + (1-a)\nu_2$ is fully-supported but not ergodic for any $0<a<1$.
–
Vaughn ClimenhagaNov 22 '12 at 2:28

I believe answer to Q1 is yes' and can be proved by point- set topological arguments. Pf that Erg. implies top. transitive. By contraposition. Say $\Phi$ is not top. transitive. Then there exist open U, V which are disj., nonempty, and invariant. Since$\mu(U) > 0$ and $\mu(V) > 0$ this proves $\Phi$ not ergodic. Fat Cantor sets' [ sets of pos. meas having empty interior] make Top. trans. imply ergodic harder. I warrant it is true though.
–
Richard MontgomeryNov 22 '12 at 3:22

Yes, existence of a fully-supported ergodic measure implies topological transitivity, by the proof you describe. But the other direction has all the problems described in the answers and comments.
–
Vaughn ClimenhagaNov 22 '12 at 3:39

4 Answers
4

For concreteness let $M=\{0,1\}^\mathbb{Z}$ be the set of bi-infinite sequences of $0$s and $1$s, and let $\Phi\colon M\to M$ be the shift map given by $\Phi(x)_j = x_{j+1}$ for $x=(x_j)_{j\in\mathbb{Z}}$.

Q1. Topological transitivity of $\Phi$ only depends on $\Phi$ and $M$, not on the measure $\mu$. In particular the system $(M,\Phi)$ defined above is topologically transitive, but there are many (many!) regular Borel probability measures that are preserved by $\Phi$, and not all of them are ergodic. See this question for some discussion of how intricate this space is. In particular, let $p$ and $q$ be fixed points for $\Phi$, and let $\mu$ be the atomic measure that gives weight $\frac 12$ to each of $p$ and $q$. Then $\mu$ is $\Phi$-invariant but not ergodic.

Q2. The pointwise time averages do not need to exist for every $x$. In fact it is quite typical that they do not exist. Let me make this last statement a little more precise, again using the example of $(M,\Phi)$ from above.

Consider the continuous real valued function $f\colon M\to \mathbb{R}$ defined by $f(x) = x_0$. That is, $f$ is simply the value of the symbol in the $0$ position in the sequence $x$. Then $a_N(x) := \frac 1N \sum_{j=1}^N f(\Phi^j(x))$ is the frequency of the symbol $1$ in the string $x_1 x_2 \cdots x_N$.

The pointwise time averages of $f$ along the orbit of $x$ exist if and only if $a_N(x)$ converges as $N\to \infty$ -- in other words, if and only if the lower and upper asymptotic frequencies of the symbol $1$ are equal. It is straightforward to construct examples of sequences $x\in M$ such that the lower and upper asymptotic frequencies disagree and the limit does not exist.

In fact, one can say some more about how large the set of such points are. Given $x\in M$, let $\lambda(x) = \liminf a_N(x)$ and $\Lambda(x) = \limsup a_N(x)$. Note that $0\leq \lambda(x)\leq \Lambda(x)\leq 1$ for all $x\in M$. Given $0\leq r\leq s\leq 1$, let $K_{r,s}$ be the set of $x\in M$ such that $\lambda(x) = r$ and $\Lambda(x) = s$. The study of the various sets $K_{r,s}$ is called multifractal analysis, and quite a lot is known. I'll state just a few results addressing your question.

Let $K^\neq = \bigcup_{r<s} K_{r,s}$ be the set of points for which $\lambda(x) \neq \Lambda(x)$, so that the limit doesn't exist. Then the following are true (at least for the system I described above -- determining for which general classes of systems these statements hold is a more subtle question):

$K^\neq$ has zero measure for every $\Phi$-invariant measure.

$K^\neq$ has Hausdorff dimension equal to the Hausdorff dimension of $M$. (The more honest way of saying this is that they have equal topological entropies, but Hausdorff dimension is a more familiar concept and the statement with dimension is true if you use an appropriate metric.)

$K^\neq$ is residual -- that is, it is a countable intersection of open and dense subsets of $M$.

In fact, one can show that $K_{0,1}$ is residual, but that it has Hausdorff dimension $0$. (The fact that $K^\neq$ has full Hausdorff dimension is due to the fact that the Hausdorff dimension of $K_{r,s}$ approaches the Hausdorff dimension of $M$ as $r,s\to \frac 12$.)

So there's an assortment of facts for you illustrating how large the set of points is where convergence fails. In particular the last fact can be interpreted as saying that from a topological point of view, for a generic point $x$ the limit fails to exist as strongly as it can possibly fail. This highlights the fact that ergodic theory is really about measures, not topology. (I will note that the limit exists everywhere if your map is uniquely ergodic, that is, if there is only one invariant probability measure. Such systems are quite different from the systems I was describing, which should be thought of as hyperbolic, or informally, chaotic.)

As pointed out in the other answers, the answer to the question Q2 is negative in general. But, there are some very special dynamical systems which are not uniquely ergodic but for which the answer to Q2 is positive, i.e. for any continuous $f:X \to \mathbb{R}$, the ergodic average

exists for all $x$, and also $f^*(x)$ depends non-trivially on $x$. The best known example of such a phenomenon is the celebrated theorem of Ratner on unipotent flows. This theorem has many applications in various fields, in particular to number theory. In all the applications, the key point is exactly that one understands all orbits (and not just almost all).

In other words, if instead of the geodesic flow on a surface of constant negative curvature with cusps you consider the horocycle flow, then the answer to Q2 is yes. (This is a theorem of Dani, which is a special case of Ratner's theorem).

There is a difference between ergodicity and unique ergodicity. Notice that for the geodesic flow space averages equal time averages for almost all, but definitely not for all orbits (as shown by the existence of closed geodesics). On the other hand the horocycle flow IS uniquely ergodic (on a compact surface). That doesn't quite answer your Q2 (the average may exist, just not equal the space average).

In any case, the best (in my opinion) exposition of the ergodicity of the geodesic flow is given by Curt McMullen in his dynamics course notes.

No - the answer to both questions is "no"; for Q1 even stronger, there are minimal transformations which are not uniquely ergodic. The corresponding examples (somewhat more complicated than plain geometric flows though) are contained in old Furstenberg's paper "Strict ergodicity and transformation of the torus" (Amer. J. Math. 83 1961).

Hmm. It seems my question Q2 is being mis-interpreted. I fix the measure. I am NOT asking if the time average equals the space average. I am asking if the time average EXISTS for all x. The Birkhoff ergodic theorem asserts that it exists for a.e. x. regardless of whether or not the map is ergodic!
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Richard MontgomeryNov 22 '12 at 3:21

No - I understood it precisely this way. Have a look at Vaughn's answer - he explains Q2 very cleary.
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R WNov 22 '12 at 8:03